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sergejj [24]
2 years ago
7

A secondary step in the process to produce ultra-pure silicon is to combine silicon tetrachloride with magnesium. How many grams

of Si could be produced by reacting 2.00 kg of SiCl4 with excess Mg
Chemistry
1 answer:
mariarad [96]2 years ago
8 0

The mass of silicon, Si produced from the reaction is 329.41 g

<h3>Balanced equation </h3>

SiCl₄ + 2Mg —> 2MgCl₂ + Si

Molar mass of SiCl₄ = 28 + (35.5×4) = 170 g/mol

Mass of SiCl₄ from the balanced equation = 1 × 170 = 170 g

Molar mass of Si = 28 g/mol

Mass of Si from the balanced equation = 1 × 28 = 28 g

From the balanced equation above,

170 g of SiCl₄ reacted to produce 28 g of Si.

<h3>How to determine the mass of Si produced </h3>

From the balanced equation above,

170 g of SiCl₄ reacted to produce 28 g of Si.

Therefore,

2 Kg (i.e 2000 g) of SiCl₄ will react to produce = (2000 × 28) / 170 = 329.41 g of Si

Thus, 329.41 g of Si were obtained from the reaction

Learn more about stoichiometry:

brainly.com/question/14735801

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Sodium peroxide (Na2O2) is often used in self-contained breathing devices, such as those used in fire emergencies, because it re
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Answer:

725.15 L

Explanation:

The balanced chemical equation for the reaction between Na₂O₂ and CO₂ is the following:

Na₂O₂ + CO₂ → Na₂CO₃ + 1/2 O₂

From the stoichiometry of the reaction, 1 mol of Na₂O₂ reacts with 1 mol CO₂. So, the stoichiometric ratio is 1 mol Na₂O₂/1 CO₂.

Now, we convert the mass of reactants to moles by using the molecular weight (Mw) of each compound:

Mw (Na₂O₂) = (23 g/mol x 2) + (16 g/mol x 2) = 78 g/mol

moles Na₂O₂ = 96.7 g/(78 g/mol) = 1.24 mol Na₂O₂

Mw (CO₂) = 12 g/mol + (16 g/mol x 2) = 44 g/mol

moles CO₂ = 0.0755 g/(44 g/mol) = 1.71 x 10⁻³ mol CO₂

Now, we calculate the number of moles of CO₂ we need to completely react with the mass of Na₂O₂ we have:

1.24 mol Na₂O₂ x 1 mol CO₂/1 mol Na₂O₂ = 1.24 mol CO₂

In 1 L of respired air we have 1.71 x 10⁻³ mol CO₂ (0.0755 g), so we need the following number of liters to have 1.24 mol of CO₂:

1.24 mol CO₂ x 1 L/1.71 x 10⁻³ mol CO₂ = 725.15 L

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